Primitive of x squared over x squared minus a squared/Logarithm Form

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Theorem

$\displaystyle \int \frac {x^2 \ \mathrm d x} {x^2 - a^2} = x + \frac a 2 \ln \left({\frac {x - a} {x + a} }\right) + C$

for $x^2 > a^2$.


Proof

Let:

\(\displaystyle \int \frac {x^2 \ \mathrm d x} {x^2 - a^2}\) \(=\) \(\displaystyle \int \frac {x^2 - a^2 + a^2} {x^2 - a^2} \ \mathrm d x\)
\(\displaystyle \) \(=\) \(\displaystyle \int \frac {x^2 - a^2} {x^2 - a^2} \ \mathrm d x + \int \frac {a^2} {x^2 - a^2} \ \mathrm d x\)
\(\displaystyle \) \(=\) \(\displaystyle \int \mathrm d x + a^2 \int \frac {\mathrm d x} {x^2 - a^2}\) Primitive of Constant Multiple of Function
\(\displaystyle \) \(=\) \(\displaystyle x + a^2 \int \frac {\mathrm d x} {x^2 - a^2} + C\) Primitive of Constant
\(\displaystyle \) \(=\) \(\displaystyle x + a^2 \left({\frac 1 {2 a} \ln \left({\frac {x - a} {x + a} }\right)}\right) + C\) Primitive of Reciprocal of $x^2 - a^2$: Logarithm Form
\(\displaystyle \) \(=\) \(\displaystyle x + \frac a 2 \ln \left({\frac {x - a} {x + a} }\right) + C\) simplifying

$\blacksquare$


Sources